Question: Simplify the following expression and state the condition under which the simplification is valid. $q = \dfrac{-3y^3 + 36y^2 - 81y}{y^3 - 5y^2 + 6y}$
First factor out the greatest common factors in the numerator and in the denominator. $ q = \dfrac {-3y(y^2 - 12y + 27)} {y(y^2 - 5y + 6)} $ $ q = -\dfrac{3y}{y} \cdot \dfrac{y^2 - 12y + 27}{y^2 - 5y + 6} $ Simplify: $ q = - 3 \cdot \dfrac{y^2 - 12y + 27}{y^2 - 5y + 6}$ Since we are dividing by $y$ , we must remember that $y \neq 0$ Next factor the numerator and denominator. $ q = - 3 \cdot \dfrac{(y - 3)(y - 9)}{(y - 3)(y - 2)}$ Assuming $y \neq 3$ , we can cancel the $y - 3$ $ q = - 3 \cdot \dfrac{y - 9}{y - 2}$ Therefore: $ q = \dfrac{ -3(y - 9)}{ y - 2 }$, $y \neq 3$, $y \neq 0$